$\lim_{x \to 0} \frac {(x^2-\sin x^2) }{ (e^ {x^2}+ e^ {-x^2} -2)} $ solution? I recently took an math exam where I had this limit to solve
$$ \lim_{x \to 0} \frac {(x^2-\sin x^2) }{ (e^ {x^2}+ e^ {-x^2} -2)}  $$
and I tought I did it right, since I proceeded like this:
1st I applied Taylor expansion of the terms to the second grade of Taylor, but since I found out the grade in the numerator and in the denominator weren't alike, I chose to try and scale down one grade of Taylor, and I found my self like this:
$$\frac{(x^2-x^2+o(x^2) )}{( (1+x^2)+(1-x^2)-2+o(x^2) )}$$
which should be:
$$\frac{0+o(x^2)}{0+o(x^2)}$$
which should lead to $0$.
Well, my teacher valued this wrong, and I think i'm missing something, I either don't understand how to apply Taylor the right way, or my teacher did a mis-correction (I never was able to see where my teacher said I was wrong, so that's why I'm asking you guys)
Can someone tell me if I really was wrong, and in case I was explain how I should have solved this?
Thanks a lot.
 A: Your working is correct, except the $o(x^2)$ should be $o(x^6)$ in the numerator and $o(x^4)$ in the denominator. But the main point is that as $x \to 0$, you get $\frac{0}{0}$.
This, however, does not equal to $0$. It's "indeterminate" or not well-defined.
You can consider proceeding by using L'Hopital's Rule, which states that in such cases, the limit does not change if we take derivatives of both numerator and denominator, and still let $x \to 0$. 
You may have to use L'Hopital's Rule more than once in some cases, or as already pointed out, factor the expression appropriately before taking derivatives to make the process easier.
A: $$\lim_{x^2\to0}\frac{x^2-\sin x^2}{e^{x^2}+e^{-x^2}-2}$$
$$=\lim_{y\to0}\frac{y-\sin y}{e^y+e^{-y}-2}$$
$$=\lim_{y\to0}\frac{y-(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots)}{(1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+\cdots)+(1-y+\frac{y^2}{2!}-\frac{y^3}{3!}+\cdots)-2}$$
$$=\lim_{y\to0}\frac{\frac{y^3}{3!}-\frac{y^5}{5!}+\cdots}{2(\frac{y^2}{2!}+\frac{y^4}{4!}+\cdots)}$$
$$=\lim_{y\to0}\frac{\frac{y}{3!}-\frac{y^3}{5!}+\cdots}{2(\frac{1}{2!}+\frac{y^2}{4!}+\cdots)}$$ Dividing the numerator & the denominator by $y^2$ as $y\ne 0$ as $y\to0$
So, $$\lim_{y\to0}\frac{y-\sin y}{e^y+e^{-y}-2}=0$$
A: $\frac{0}{0}$ is indeterminate-- not $0$. Notice you can factor the bottom (which may make the next step easier):
Apply L'Hospital twice.
A: How is $\frac{0+o(x^2)}{0+o(x^2)}$ zero? 
You need to expand to a degree high enough to keep something nontrivial after cancellation!
Note that $\sin(x^2)=x^2-\frac12 x^4+o(x^6)$ and
$e^{\pm x^2}=1+\pm x^2+\frac 12 x^4+o(x^6)$, hence
$$f(x)= \frac{\frac12 x^4 + o(x^6)}{x^4+o(x^6)}=\frac{\frac12  + o(x^2)}{1+o(x^2)}\to \frac 12$$
A: It can also be written as ${x^2-\sin x^2}\over{4\sinh^2{x^2/2}} $ $\approx {{x^2-(x^2-x^6/3!)}\over{x^4}}$ $=1/6$
A: Re-writing the original expression, the result follows easily
\begin{equation*}
\lim_{x\rightarrow 0}\frac{x^{2}-\sin (x^{2})}{e^{x^{2}}+e^{-x^{2}}-2}%
=\lim_{t\rightarrow 0}\frac{t-\sin (t)}{e^{t}+e^{-t}-2}=\lim_{t\rightarrow
0}\left( \frac{t-\sin t}{t^{3}}\right) \frac{t}{\left( \frac{e^{t}-1-t}{t^{2}%
}\right) +\left( \frac{e^{-t}-1+t}{t^{2}}\right) }=\left( \frac{1}{6}\right) 
\frac{0}{\left( \frac{1}{2}\right) +\left( \frac{1}{2}\right) }=0.
\end{equation*}
The key idea in the transformation is to get known expressions. The limits of which can be obtained by L'hospital's rule successively 2 or 3 times.
